# Prime ideal generated by $\operatorname{height}P$ elements

Let $P$ be a prime ideal of height $n$ in a local ring $R$, which is generated by $n$ elements, say $P=(a_1,...,a_n)$. The image $\bar P$ of $P$ in $R/(a_1,...,a_i)$ , $1≤i≤n$, is an $(n-i)$-generated prime. I want to be sure that it is of height $n-i$.

The height is at most $n-i$, because the height of $\bar P$ is less than or equal to the height of any minimal prime over it in $R/(a_1,...,a_i)$, and the latter is at most $n-i$ by Principal Ideal Theorem. Now, why is it at least $n-i$?

• @user26857 For $i=1$, the height of $P/(a_1)$ equals $n$ or $n-1$, so at least $n-1=n-i$. Could this be used in general? – karparvar Sep 9 '15 at 9:29
• Not really.${}$ – user26857 Sep 9 '15 at 9:32
• @user26857 Would you please give your elementary solution? I really have known that $a_1,...,a_n$ is an $R$-sequence, so it is part of an s.o.p., hence $dim (a_1,...,a_i)=d-i$, where $d$ is dim of $R$.... – karparvar Sep 9 '15 at 9:40
First, I think, it is not necessary to assume that $R$ is local. Instead consider the localization $S=R_P$ with maximal ideal $M = P R_P$. Then, as height $P$ is $n$ and $P$ is generated by $n$ elements, the ring $S$ is a regular local ring. As $M = (a_1,\ldots,a_n)$ the sequence $a_1,\ldots,a_n$ is a regular sequence in $S$. So for $I=(a_1,\ldots,a_i)$ we have $\dim S/I = \operatorname{depth} S/I = n - i$, because $a_{i+1} + I,\ldots,a_n + I$ is a regular sequence in $S/I$. As $M/I$ is generated by the $n-i$ elements $(a_{i+1},\ldots,a_n)$, we have that $S/I$ is a regular local ring, therefore integral domain, and $I$ is a prime (in $S$). So, as $\dim S/I = n - i$ there exists a chain of prime ideals
$$I = P_0 \subseteq \cdots \subseteq P_{n-i} = M$$
in $S$. Considering the $P_i$ as prime ideals in $R$ gives your sought after sequence of length $n - i$.